Lower bound for conditional probability

Question

I have $X_1,X_2$ both identically and independently distributed $\text{Bin}(n,\theta)$. For some $\theta_0\in(0,1)$, and integers (depending on $n$) $a_n$ and $b_n$ satisfying $$ n\theta_0\leq a_n< n\\ 2n\theta_0\leq b_n< 2a_n\\ $$ and $a_n\to\infty$, $b_n\to\infty$ as $n\to\infty$. I want to show that $$ P(X_1+X_2>b_n\mid X_1\leq a_n,X_2\leq a_n,\theta\leq\theta_0)\geq P(X_1+X_2>b_n\mid \theta\leq\theta_0) $$ Intuitively I think this is true since $$ P(X_1\leq a_n,X_2\leq a_n\mid \theta\leq\theta_0)\geq P(X_1\leq a_n,X_2\leq a_n\mid \theta_0)\to 1 $$ as $n\to\infty$. However, I can't seem to prove the inequality or show an counterexample that it's not true.

Answer 1

Let $a_n = n\theta_0$. Then $X_1\le a_n$ and $X_2\le a_n$ implies $X_1+X_2 \le 2a_n = 2n\theta_0 \le b_n$. This makes $$P(X_1+X_2 > b_n | X_1 \le a_n, X_2 \le a_n, \theta \le \theta_0) = 0$$ Since $$ P(X_1+X_2 > b_n | \theta \le \theta_0) \ne 0$$ the inequality does not hold.